How do you solve #x^2=-9x-8#? Algebra Quadratic Equations and Functions Comparing Methods for Solving Quadratics 1 Answer Hgsfjc S. Apr 9, 2018 x= -1 or-8 Explanation: First use null factor law to put the #-x^2# on the other side hence making 0= #-x^2# -9x-8. Simplify by dividing by-1 = #x^2# +9x+8=0. Then use cross method to factorise which becomes (x+8)(x+1)=0 the x being -8 or -1 as the answer Answer link Related questions What are the different methods for solving quadratic equations? What would be the best method to solve #-3x^2+12x+1=0#? How do you solve #-4x^2+4x=9#? What are the two numbers if the product of two consecutive integers is 72? Which method do you use to solve the quadratic equation #81x^2+1=0#? How do you solve #-4x^2+4000x=0#? How do you solve for x in #x^2-6x+4=0#? How do you solve #x^2-6x-16=0# by factoring? How do you solve by factoring and using the principle of zero products #x^2 + 7x + 6 = 0#? How do you solve #x^2=2x#? See all questions in Comparing Methods for Solving Quadratics Impact of this question 2319 views around the world You can reuse this answer Creative Commons License